If [t] denotes the greatest integer <math><mrow><mo>≤</mo><mi>t</mi><mo>,</mo></mrow></math> then the value of <math><mrow><mstyle displaystyle="true"><mrow><msubsup><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></msubsup><mrow><mrow><mo>[</mo><mrow><mn>2</mn><mi>x</mi><mo>−</mo><mrow><mo>|</mo><mrow><mn>3</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>2</mn></mrow><mo>|</mo></mrow><mo>+</mo><mn>1</mn></mrow><mo>]</mo></mrow><mi>d</mi><mi>x</mi><mtext> </mtext><mtext> </mtext><mi>i</mi><mi>s</mi><mo>:</mo></mrow></mrow></mstyle></mrow></math>

y = 2x |3x2−5x+2|,0≤x≤1={−3x2+7x−2,3x2−3x+2,0≤x≤2323<x≤13x2 – 5x + 2 = 0x=+5±25−246= 5+16=1,233x2 – 7x + 3 = 0x = 7±49−396=7±136−3x2+7x−2−7±49−24−6=7+56=2,133x2 + 7x – 2 = 13x2 – 7x + 1 = 0x = 7±49−126=7±376I = ∫06((−2)+1)dx+∫7−37613((−1)+1)dx+∫−137−136(0+1)dx+∫7−13623(1+1)dx+∫231(1+1)dx=37+13−46

If [t] denotes the greatest integer&nbsp;<math><mrow><mo>≤</mo><mi>t</mi><mo>,</mo></mrow></math>&nbsp;then the value of&nbsp;<math><mrow><mstyle displaystyle="true"><mrow><msubsup><mo>∫</mo><mrow><mn>0</mn></mrow><mrow><mn>1</mn></mrow></msubsup><mrow><mrow><mo>[</mo><mrow><mn>2</mn><mi>x</mi><mo>−</mo><mrow><mo>|</mo><mrow><mn>3</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>2</mn></mrow><mo>|</mo></mrow><mo>+</mo><mn>1</mn></mrow><mo>]</mo></mrow><mi>d</mi><mi>x</mi><mtext> </mtext><mtext> </mtext><mi>i</mi><mi>s</mi><mo>:</mo></mrow></mrow></mstyle></mrow></math>

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Greatest Integer Function Relations and Functions Maths Class 11th

If [t] denotes the greatest integer $\leq t,$ then the value of $\int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x i s :$

Option 1 -

$\frac{\sqrt[]{37} + \sqrt[]{13} - 4}{6}$

Option 2 -

$\frac{\sqrt[]{37} - \sqrt[]{13} - 4}{6}$

Option 3 -

$\frac{- \sqrt[]{37} - \sqrt[]{13} + 4}{6}$

Option 4 -

$\frac{- \sqrt[]{37} + \sqrt[]{13} + 4}{6}$

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

y = 2x $| 3 x^{2} - 5 x + 2 |, 0 \leq x \leq 1$

$= {\begin{matrix} - 3 x^{2} + 7 x - 2, & 3 x^{2} - 3 x + 2, \end{matrix} \begin{matrix} 0 \leq x \leq \frac{2}{3} & \frac{2}{3} < x \leq 1 \end{matrix}$

3x²– 5x + 2 = 0

$x = \frac{+ 5 \pm \sqrt[]{25 - 24}}{6}$

= $\frac{5 + 1}{6} = 1, \frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7 \pm \sqrt[]{49 - 39}}{6} = \frac{7 \pm \sqrt[]{13}}{6}$

$- 3 x^{2} + 7 x - 2$

$\frac{- 7 \pm \sqrt[]{49 - 24}}{- 6} = \frac{7 + 5}{6} = 2, \frac{1}{3}$

3x² + 7x – 2 = 1

3x² – 7x + 1 = 0

x = $\frac{7 \pm \sqrt[]{49 - 12}}{6} = \frac{7 \pm \sqrt[]{37}}{6}$

I = $\int_{0}^{6} (\begin{matrix} (- 2) & + 1 \end{matrix}) d x + \int_{\frac{7 - \sqrt[]{37}}{6}}^{\frac{1}{3}} (\begin{matrix} (- 1) & + 1 \end{matrix}) d x + \int_{- \frac{1}{3}}^{\frac{7 - \sqrt[]{13}}{6}} (\begin{matrix} 0 & + 1 \end{matrix}) d x + \int_{\frac{7 - \sqrt[]{13}}{6}}^{\frac{2}{3}} (1 + 1) d x + \int_{\frac{2}{3}}^{1} (1 + 1) d x$

$= \frac{\sqrt[]{37} + \sqrt[]{13} - 4}{6}$

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If [t] denotes the greatest integer $\leq t,$ then the value of $\int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x i s :$

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If [t] denotes the greatest integer $\leq t,$ then the value of $\int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x i s :$